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Evaluate $$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} \text{d}x$$
where $n\in\mathbb{N}$

I was able to do this using parametization (differentiation under integration) but I'm not allowed to use that in my school since it isn't in the syllabus :(
I should only use high school level mathematics. However, I can't see a way to do this with an elementary approach.

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  • $\begingroup$ @Integrator In that question the limits are different $\endgroup$ – Henry Dec 2 '14 at 18:38
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Hint:

Start by using the power reduction formula to get rid of the sine-squared term:

$$\begin{align} \mathcal{I}_{n} &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{nx}}{\sin{x}}\,\mathrm{d}x\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{2nx}}{2\sin{x}}\,\mathrm{d}x.\\ \end{align}$$

Then, writing the $\cos{2nx}$ as a binomial series, we can rewrite the integrand as a polynomial of sines and cosines:

$$\cos{(2nx)}=\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=1-\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=1-\cos^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies 1-\cos{(2nx)}=\sin^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$

$$\implies \frac{1-\cos{(2nx)}}{\sin{(x)}}=\sin{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k-1}{(x)}.$$

Finally, integrate term-by-term.

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  • $\begingroup$ In the last step, why is the interchanging of summation and integral allowed? $\endgroup$ – Henry Dec 2 '14 at 18:52
  • $\begingroup$ So, will we use beta function here to evaluate the latter expression? $\endgroup$ – Venus Dec 2 '14 at 18:59
  • $\begingroup$ @Samurai Because integration is a linear operation, and the sum has only a finite number of terms. You can always swap the order for finite summations because no limits are involved. $\endgroup$ – David H Dec 2 '14 at 18:59
  • $\begingroup$ @DavidH Thanks. $\endgroup$ – Henry Dec 2 '14 at 19:00
  • $\begingroup$ @Venus The OP specified that only high school methods should be employed, so wouldn't consider beta function machinery appropriate here. You certainly don't need it. All can be accomplished with basic finite binomial sums. $\endgroup$ – David H Dec 2 '14 at 19:05
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Recall the identity: $$\sin^2(A) - \sin^2(B) = \sin(A+B)\sin(A-B)$$ Let $I_n = \displaystyle \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx$, we obtain \begin{align} I_{n+1} - I_n & = \int_0^{\pi/2} \dfrac{\sin^2((n+1)x)}{\sin(x)}dx - \int_0^{\pi/2} \dfrac{\sin^2(nx)}{\sin(x)}dx\\ & = \int_0^{\pi/2}\dfrac{\sin^2((n+1)x)-\sin^2(nx)}{\sin(x)} dx\\ & = \int_0^{\pi/2}\dfrac{\sin((2n+1)x)\sin(x)}{\sin(x)} dx\\ & = \int_0^{\pi/2} \sin((2n+1)x)dx\\ & = -\left.\dfrac{\cos((2n+1)x)}{2n+1} \right \vert_0^{\pi/2}\\ & = \dfrac1{2n+1} \end{align} We have $I_0 = 0$ and hence $$I_{n} = \dfrac11 + \dfrac13 + \dfrac15 + \cdots + \dfrac1{2n-1}$$

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